There is a map (i.e. a commutative diagram) from the exact sequence $$0\to Ext(H_{n-1}(A),G)\to H^n(A;G)\to Hom(H_n(A),G)\to 0$$ to the exact sequence $$0\to Ext(H_n(X,A),G)\to H^{n+1}(X,A;G)\to Hom(H_{n+1}(X,A),G)\to 0$$All the groups are functors of both the pair $(X,A)$ and the abelian group $G$. All the maps are natural. The exact sequences split, but the splittings are not natural, so there is not a canonical map $Hom(H_n(A),G)\to Ext(H_n(X,A),G)$ to inquire about. However, by the commutative diagram there is a canonical map from a subgroup of the former to a quotient of the latter, namely from $$ker(Hom(H_n(A),G)\to Hom(H_{n+1}(X,A),G)=Hom(coker(H_{n+1}(X,A)\to H_n(A)),G)=Hom(P,G)$$ to $$coker(Ext(H_{n-1}(A),G)\to Ext(H_n(X,A),G))=Ext(ker(H_n(X,A)\to H_{n-1}(A)),G)=Ext(Q,G)$$ where $P=Im(H_n(A)\to H_n(X))$ and $Q=H_n(X)/P$. This is in fact determined by the exact sequence $0\to P\to H_n(X)\to Q\to 0$, part of the six-term exact sequence $$0\to Hom(Q,G)\to Hom(H_n(X),G)\to Hom(P,G)\to Ext(Q,G)\to Ext(H_n(X),G)\to Ext(P,G)\to 0$$
This map can easily be nonzero when $G=\mathbb Z/m$, for example if $H_{n+1}(X,a)\to H_n(A)\to H_n(X)\to H_n(X,A)\to H_{n-1}(A)$ is the exact sequence $0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\to 0$ and $p$ divides $m$.
I'll start by describing the notation that I'll use.
I'll think of elements of $\mathbb{Z}^\mathbb{N}$ as infinite column vectors
$${\bf x}=\begin{pmatrix}
x_0\\x_1\\x_2\\\vdots
\end{pmatrix}$$
of integers, with rows indexed by $\mathbb{N}$.
The $i$th "unit vector" will be denoted by ${\bf e(i)}$. I.e., $e(i)_i=1$ and $e(i)_j=0$ for $j\neq i$.
As described in the question, Specker's result implies that endomorphisms of $\mathbb{Z}^\mathbb{N}$ are of the form ${\bf x}\mapsto A{\bf x}$ where
$$A=\begin{pmatrix}
a_{00}&a_{01}&a_{02}&\cdots\\
a_{10}&a_{11}&a_{12}&\cdots\\
a_{20}&a_{21}&a_{22}&\cdots\\
\vdots&\vdots&\vdots&\ddots
\end{pmatrix}$$
is an infinite matrix of integers, with rows and columns indexed by $\mathbb{N}$, that is row-finite (i.e., each row has only finitely many non-zero entries).
Now suppose that $\alpha:\mathbb{Z}^\mathbb{N}\to\mathbb{Z}^\mathbb{N}$ is a surjective group homomorphism, given by a row-finite matrix $A$.
To prove that $\alpha$ is split, we need to find a right inverse. In fact, it suffices to find another row-finite matrix $B$ such that $AB$ is lower unitriangular (i.e., diagonal entries are all $1$ and entries above the diagonal are all $0$), since such a matrix is invertible.
Fix $n\geq0$, and decompose $\mathbb{Z}^\mathbb{N}=G_n\oplus H_n$, where $G_n\cong\mathbb{Z}^n$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i\geq n$, and $H_n\cong\mathbb{Z}^\mathbb{N}$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i<n$.
Since $\alpha$ is surjective, $\mathbb{Z}^\mathbb{N}=\alpha\left(G_n\right)+\alpha\left(H_n\right)$.
Consider the quotient maps
$$\mathbb{Z}^\mathbb{N}\stackrel{\theta}{\to} \mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\stackrel{\varphi}{\to} \left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)/T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right),$$
where $T(X)$ denotes the torsion subgroup of an abelian group $X$.
Since $\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)/T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$ is a finite rank free abelian group, using Specker's result again implies that $\varphi\theta({\bf e(i)})=0$, or equivalently $\theta({\bf e(i)})\in T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$, for all but finitely many $i$.
Since $T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$ is finite, for all but finitely many $i$ with $\theta({\bf e(i)})\in T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$, there is $i'>i$ with $\theta({\bf e(i')})=\theta({\bf e(i)})$, or equivalently ${\bf e(i)}-{\bf e(i')}\in\alpha(H_n)$.
So we can choose a sequence $0=t_0<t_1<t_2<\dots$ of integers such that for every $i\geq t_n$ there is some $i'>i$ and some ${\bf b(i)}\in H(n)$ with ${\bf e(i)}-{\bf e(i')}=\alpha\left({\bf b(i)}\right)$. For each $i$, do this with the largest $n$ such that $i\geq t_n$, and let $B$ be the matrix whose $i$th column is ${\bf b(i)}$.
Then $B$ is row-finite, and $AB$ is lower unitriangular.
Best Answer
Consider the classifying space $EG$ of a given group $G$. In your case, $G={Spin}(n)$. We have a fibration $G\to EG\to BG$ where $EG$ is contractible and $G$ acts freely on it. Therefore, the long exact sequence in homotopy groups tells you that $\pi_j(BG)\cong \pi_{j-1}(G)$. But $G={Spin}(n)$ is a Lie group which is simply connected. It is also classically known that $\pi_2(G)=0$ for finite dimensional Lie groups $G$. And, we also know that $\pi_3({Spin}(n))=\mathbb{Z}$. This implies that $\pi_j(B {Spin}(n))=0$ for $j=0,1,2,3$. Moreover, $\pi_4(B {Spin}(n))\cong H_4(B{Spin}(n);\mathbb{Z})$ by Hurewicz theorem. This is also isomorphic to $\pi_3({Spin}(n))$. The universal coefficient theorem now tells you that $H^4(B{Spin}(n);\mathbb{Z})=\mathbb{Z}$.